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Mechanical Properties Of Fluids

Question
CBSEENPH11020786

A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then:

  • straight T proportional to square root of straight rho
  • straight T proportional to fraction numerator 1 over denominator square root of straight A end fraction
  • straight T proportional to 1 over straight rho
  • straight T proportional to fraction numerator 1 over denominator square root of straight m end fraction

Solution

B.

straight T proportional to fraction numerator 1 over denominator square root of straight A end fraction

Force applied on the body will be equal to upthrust for vertical oscillations.
Let block is displaced through x m, then weight of displaced water or upthrust (upwards)
    equals negative Axρg
where A is area of cross-section of the block and straight rho is its density. This must be equal to force (=ma) applied, where m is mass of the block and straight alpha is acceleration.
therefore space space space space ma space equals space minus Axρg
 or                straight alpha space equals space minus Aρg over straight m straight x space equals space minus straight omega squared straight x
This is the equation of simple harmonic motion.
Time period of oscillation
              straight T equals fraction numerator 2 straight pi over denominator straight omega end fraction equals 2 straight pi square root of straight m over Aρg end root
rightwards double arrow space space space space space straight T proportional to fraction numerator 1 over denominator square root of straight A end fraction